Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=-2x\)
Đặt \(\hept{\begin{cases}\sqrt{4+x}=a\ge0\\\sqrt{4-x}=b\ge0\end{cases}}\) thì ta có:
\(\hept{\begin{cases}\left(a-2\right)\left(b+2\right)=b^2-a^2\left(1\right)\\8=a^2+b^2\left(2\right)\end{cases}}\)
Lấy (2) + 2.(1) vế theo vế rút gọn ta được
\(\Leftrightarrow3b^2-a^2+4b-4a-2ab=0\)
\(\Leftrightarrow\left(b-a\right)\left(3b+a+4\right)=0\)
\(\Leftrightarrow a=b\)
\(\Rightarrow\sqrt{4+x}=\sqrt{4-x}\)
\(\Leftrightarrow x=0\)
Ta có : \(\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)=-2x\)
\(\Rightarrow\left(\sqrt{x+4}\right)^2-2^2=-2x\)
\(\Leftrightarrow x+4-4=-2x\)
=> x = -2x
=> x + 2x = 0
=> 3x = 0
=> x = 0
Vậy x = 0.
Câu a : \(3\sqrt{x-2}-\sqrt{x^2-4}=0\) ( ĐK : \(x\ge2\) )
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=7\left(TM\right)\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=7\)
1) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
2) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
3) \(\left(x+\frac{1}{3}\right)^4=\left[\left(x+\frac{1}{3}\right)^2\right]^2=\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)^2=x^4+\frac{4}{9}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x+\frac{2}{9}x^2=x^4+\frac{2}{3}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x\)
4) \(\left(2x+y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
5) Sửa đề: \(\left(\frac{x}{2}-2y\right)^3=\frac{x^3}{8}-\frac{3x^2}{2}+6xy^2-8y^3\)
6) \(\left(\sqrt{2x-y}\right)^4=\left(2x-y\right)^2=4x^2-4xy+y^2\)
7) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
8) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)